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					  Redox Reactions

oxidation reduction reactions
Ch 22 sec 1
From the combustion of gasoline to the
 metabolism of food - oxidation is
 responsible
Oxidation - Loss of electrons
  old definition - gain of oxygen
Reduction - Gain of electrons
  old definition - loss of oxygen
ONE DOES NOT OCCUR WITHOUT THE
 OTHER
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           LEO the lion goes GER
Losing Electrons is Oxidation
Gaining Electrons is Reduction
Example
  Mg + S ---> Mg2+ + S2-
Magnesium is oxidized (aka reducing
 agent)
Sulfur is reduced (aka oxidizing agent)
  Redox reactions are usually
presented as two components
Mg ----> Mg2+ + 2e-
S + 2e- ---> S2-
Identifying transfers of electrons is easy
 for ionic reactions.
What about covalent where there is not
 a transfer of electrons but a sharing?
                       Consider
            2H2 + O2 ---> 2H2O
Which element is reduced and oxidized?
 Explain
Oxygen is the electron hog. The partial
 gain of electrons means it is reduced
 and hydrogen’s partial loss means it is
 oxidized.
                       Your Turn
           4Fe + 3O2 ---> 2Fe2O3
Write the 2 redox reactions. Which is oxidized
 and which is reduced.
answer
Fe --> Fe3+ + 3e- oxidized (reducing agent)
O2 + 2e- ---> O2- reduced (oxidizing agent)
Corrosion occurs more rapidly in the presence
 of salts and acids. Why?
These are conducting solutions that make
 electron transfer easier.
 Sometimes Corrosion is good.
Aluminum oxidizes to form tightly
 packed aluminum oxide particles. This
 is a protective covering.
Iron needs to be coated to protect it
 because the oxide is not tightly packed.
                          Ch 22 sec 2
         Assigning Oxidation Numbers
Redox equations will be balanced using
 oxidation numbers
Rules
1. The oxidation number of a monatomic ion
 is equal to its ionic charge.
2. The oxidation number of hydrogen in a
 compound is +1 except in a metal hydride.
 Example - NaH H is -1
3. The oxidation number for oxygen is -2
 unless in a peroxide H2O2 where it is -1
                          more rules
The oxidation number of an
 uncombined atom (elemental form) is
 0.
For any neutral compound, the sum of
 the oxidation numbers of the atoms in
 the compound must equal 0.
For a polyatomic ion, the sum of the
 oxidation numbers must equal the ionic
 charge of the ion.
What is the oxidation number of elements
             in the following compounds?
1. SO2
S is +4 and O is -2

2. K2SO4
K is +1 S is +6 O is -2
    Oxidation numbers change in redox
         reaction. No change no redox.
An increase in the oxidation number
 indicates oxidation
A decrease in the oxidation number
 indicates reduction
Example - What is being oxidized and
 what is being reduced?
    2AgNO3 + Cu --> Cu(NO3)2 + 2Ag
     +1   +5 -2   0    +2 +5 -2        0
                              Ch 22 sec 3
                    Classifying Reactions
Either electrons are transferred or they are
 not.
Redox reactions include single-replacement,
 combination, decomposition and combustion.
Others - double replacement and acid/base
 reactions.
Color changes signify redox reactions
  video example
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                             another redox clip




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     Balancing Redox Reactions
                                Two Methods
Many reactions are too complex to be
 balanced by trial and error.
All reactions should be balanced in this
 manner. You were taught to balance atoms
 because for simple reactions it works.
 Balancing atoms may balance the equation
 and may not be the correct balanced
 equation. Atoms along with charges need to
 be balanced.
    oxidation-number-change method

balanced by comparing the increase
 and decrease of oxidation numbers
Example:
Step 1 - Assign Oxidation Numbers
 +1 +6 -2   +1-2   0   +1-2+1 +3 -2   +4-
  2
K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
 +1 +6 -2   +1-2   0   +1-2+1 +3 -2   +4-2
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2


Step 2 - Identify which atoms are oxidized
 and which are reduced.
Cr is reduced and S is oxidized
Onto step 3
Use a bracket line to connect the atoms that
 undergo oxidation and another line to
 connect those that undergo reduction. Then
 write the oxidation-number change at the
 midpoint of each line.
                         -3

 +1 +6 -2   +1-2    0        +1-2+1   +3 -2   +4-2
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
                                +4
 Step 4 - Make the total increase in oxidation number
  equal to the total decrease in oxidation number by
  using appropriate coefficients.
                 (4)(-3)=-12
 K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2
                           (3)(+4)=+12
 The coefficient is the number of atoms needed.
 2 K2Cr2O7 + H2O + 3S --> KOH + 2Cr2O3 + 3SO2
Step 5 - Finish balancing by inspection
 2K2Cr2O7 + 2H2O + 3S --> 4KOH + 2Cr2O3 + 3SO2

 Your Turn -
 Balance using the oxidation-number-change method.
As2O3 + Cl2 + H2O --> H3AsO4 + HCl
                     (2)(-1)=-2
 +3 -2       0      +1 -2      +1 +5 -2       +1-
  1
As2O3 + Cl2 + H2O --> H3AsO4 + HCl
                  (1)(+2)=+2
As2O3 + 2Cl2 +5H2O --> 2H3AsO4 + 4HCl
             Using Half-Reactions
Good for ionic reactions
Two equations used - one shows
 oxidation the other reduction.Then
 combined together in the last step.
                            Balance using 1/2 reactions
                                              Example:
 KMnO4 + HCl --> MnCl2 + Cl2 + H2O + KCl
Step 1 - Write the unbalanced equation in
 ionic form. Place oxidation numbers above.
 +1    +7-2      +1     -1       +2      -1       0   +1 -2 +1       -1
 K1+ + MnO41- + H1+ + Cl1- --> Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-
 Step 2 - Write separate 1/2 reactions for the
  oxidation reduction process.
 Cl1- --> Cl2
 MnO41- --> Mn2+
Step 3 - Balance the atoms in each 1/2
 reaction. When reactions take place in an
 acid solution you will need to use water and
 H+ to balance hydrogen and oxygen atoms.
 2Cl1- --> Cl2
 MnO41- + 8H+ --> Mn2+ + 4H2O
 atoms are balanced but charges are not.
Step 4 - Add electrons to one side of each 1/2
 reaction to balance the charges.
 2Cl1- --> Cl2 + 2e-
 MnO41- + 8H+ + 5e- --> Mn2+ + 4H2O
Step 5 - numbers of electrons must be equal.
 Multiply each reaction by a number to make
 electrons equal. In this case make electrons
 equal 10.
 10Cl1- --> 5Cl2 + 10e-
 2MnO41- + 16H+ + 10e- --> 2Mn2+ + 8H2O
Step 6 - Add the 1/2 reactions to show an
 overall equation.
 10Cl1- + 2MnO41- + 16H+ + 10e- --> 5Cl2 + 10e- + 2Mn2+ + 8H2O

Remove terms that are the same on both
 sides.
 10Cl- + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 8H2O
 Step 7 - Add spectator ions and balance the
  equation.
 From step 1
    K1+ + MnO41- + H1+ + Cl1- -->
   Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-
 10Cl- + 2K+ + 2MnO4 + 16H+ + 6Cl- --> 5Cl2 + 2Mn2+
  + 4Cl-+ 8H2O + 2K+ + 2Cl-
 Summarize spectator and nonspectator ions.
 16Cl- + 2K+ + 2MnO4 + 16H+ -->
   5Cl2 + 2Mn2+ + 6Cl- + 8H2O + 2K+
The equation is now balanced for atoms and charge.
Now you can rewrite it into:
2KMnO4 + 16HCl --> 2MnCl2 + 5Cl2 + 8H2O + 2KCl
      Balance using 1/2 reaction method
                              Your Turn:
S + HNO3 --> SO2 + NO + H2O
Step 1 - Write the unbalanced equation in
 ionic form. Place oxidation numbers above.
 0     +1   +5 -2   +4-2   +2-2   +1 -2
S + H+ + NO3- --> SO2 + NO + H2O
Identify what is oxidized and what is reduced.
Step 2 - Write separate half-reactions for the
 oxidation and reduction process.
S --> SO2
NO3- --> NO
Step 3 - Balance the atoms in each 1/2
 reaction. When reactions take place in an
 acid solution you will need to use water and
 H+ to balance hydrogen and oxygen atoms.
2H2O + S --> SO2 + 4H+
4H+ + NO3- --> NO + 2H2O
note: charges are not balanced
Step 4 - Add electrons to one side of
 each 1/2 reaction to balance the
 charges.
2H2O + S --> SO2 + 4H+ + 4e-
4H+ + NO3- + 3e- --> NO + 2H2O
Step 5 - numbers of electrons must be
 equal. Multiply each reaction by a
 number to make electrons equal. In this
 case make electrons equal 12.
6H2O + 3S --> 3SO2 + 12H+ + 12e-
16H+ + 4NO3- + 12e- --> 4NO + 8H2O
Step 6 - Add the 1/2 reactions to show
 an overall equation.
6H2O + 3S + 16H+ + 4NO3- + 12e- -->
 3SO2 + 12H+ + 12e- + 4NO + 8H2O
Remove terms that are the same on
 both sides.
3S + 4H+ + 4NO3- --> 3SO2 + 4NO + 2H2O
Step 7 - Add spectator ions and balance the
 equation. This reaction does not have
 spectator ions. Therefore final answer is:
3S + 4HNO3 --> 3SO2 + 4NO + 2H2O
  Balancing Redox Reactions in a Basic Solution

MnO4- + CN- --> MnO2 + CNO-
Initially assume acid solution but eventually
 switch over to a base.
Two 1/2 reactions are:
MnO4- --> MnO2
CN- --> CNO-
Balance as if an acid:
4H+ + MnO4- --> MnO2 + 2H2O
H2O + CN- --> CNO- + 2H+
Now reality sets in. Switch over to a base by
 adding the same number of OH- ions as H+
4OH- + 4H+ + MnO4- --> MnO2 + 2H2O + 4OH-
2OH- + H2O + CN- --> CNO- + 2H+ + 2OH-
Next, combine H+ and OH- to make water and
 cancel if possible
2 H2O + MnO4- --> MnO2 + 4OH-
2OH- + CN- --> CNO- + H2O
Mass balance has been achieved. Charge balance
 is next.
3e- + 2 H2O + MnO4- --> MnO2 + 4OH-
2OH- + CN- --> CNO- + H2O + 2e-
Electrons need to balance - multiply by LCM
6e- + 4 H2O + 2 MnO4- --> 2 MnO2 + 8OH-
6OH- + 3CN- --> 3CNO- + 3H2O + 6e-
Sum the reactions canceling like terms to give the
 net ionic equation:
H2O + 2MnO4- + 3CN- --> 2MnO2 + 2OH- + 3CNO-
            Electrochem Intro



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                                 Ch 23 sec 1
                        Electrochemical Cells
When a strip of zinc metal is placed in CuSO4.
 Electrons are transferred from Zn to Cu.
  Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
The flow of electrons is an electric current.
Any conversion between electrical and
 chemical energy is an electrochemical
 process. The device that converts between
 the two is an electrochemical cell.
                          Voltaic Cells
Convert chemical energy into electrical
 energy
Half-cell is part of the cell where
 oxidation or reduction takes place.
A half-cell consists of a metal rod or
 strip immersed in a solution of its ions.
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                 zinc-copper reaction
One 1/2 cell has a zinc rod immersed in zinc
 sulfate
One 1/2 cell has a copper rod immersed in
 copper (II) sulfate
Cells are separated by a salt bridge
  a tube containing a strong electrolyte often K2SO4
A wire carries the electrons in the external
 circuit from the zinc rod to the copper rod.
The driving force is the spontaneous redox
 reaction
Electrode - a conductor in a circuit that
 carries electrons to or from a substance.
Anode - electrode at which oxidation occurs.
 Electrons are produced at the anode so it it
 labeled the negative electrode.
Cathode - electrode at which reduction takes
 place. Electrons are consumed at the
 electrode. The cathode is labeled the positive
 electrode.
Neither electrode is really charged. The
 moving electrons balance any charge that
 might build up as oxidation and reduction
 occur.
    The electrochemical process in a zinc-copper
voltaic cell. These steps occur at the same time.

1. Electrons are produced at the zinc rod.
     Zn --> Zn2+ + 2e- (anode)
2. The electrons leave the the zinc anode and
 travel through the external circuit to the
 copper rod. If a bulb is in the circuit it will
 light.
3. Electrons enter the copper rod and interact
 with copper ions in solution.
  Cu2+ + 2e- --> Cu
To complete the circuit, both positive
 and negative ions move through the
 aqueous solutions via the salt bridge.
Summary:
  Zn --> Zn2+ + 2e-
  Cu2+ + 2e- --> Cu
  Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)
                  Dry Cells (batteries)
A voltaic cell in which the electrolyte is a
 paste.
A zinc container is filled with a thick , moist
 electrolyte paste of manganese (IV) oxide,
 (MnO2), ZnCl2, NH4Cl and water.
A graphite rod is embedded in the paste.
The zinc container is the anode and the
 graphite rod is the cathode.
The thick paste and its surrounding paper
 liner prevent the contents of the cell from
 freely mixing.
Dry Cell Reactions:
  Zn --> Zn2+ + 2e-
  2MnO2 + 2NH4+ + 2e- --> Mn2O3 + 2NH3 + H2O
The graphite serves as a conductor even
 though it is the cathode. The Mn is actually
 reduced.
Alkaline battery - same as dry cell but the
 paste is KOH to prevent buildup of ammonia
 gas.
            Lead Storage Batteries
A battery is actually a group of cells
 connected together.
A 12 volt car battery consists of 6 cells.
One set of grids packed with spongy lead
 (anode).
The other grid is (cathode) is packed with
 PbO2.
The grids are immersed in concentrated
 sulfuric acid.
The reactions:
  Pb + SO42- --> PbSO4 + 2e- (oxidation)
  PbO2 + 4H+ + SO42- + 2e- --> PbSO4 +
   2H2O (reduction)
Overall
  Pb + PbO2 + 2H2SO4 --> 2PbSO4 + 2H2O
The sulfate slowly builds up and
 concentration of acid decreases.
The car’s generator reverses the
 reaction.
                               Fuel Cells
Cells with renewable electrodes.
A voltaic cell in which a fuel substance
 undergoes oxidation and from which electrical
 energy is continuously obtained.
Do not have to be recharged
emits no air pollutants, operates quieter,
 more cost effective than an electrical
 generator.
       The hydrogen-oxygen fuel cell

3 compartments separated by 2 carbon
 electrodes.
Oxygen (the oxidizer) is fed into the
 cathode compartment.
Hydrogen (the fuel) is fed into the
 anode compartment.
The gases diffuse slowly.
The electrolyte in the center is a hot,
 concentrated solution of KOH.
Electrons enter from the oxidation 1/2
 reaction at the anode then pass through
 an external circuit to enter the
 reduction 1/2 reaction at the cathode.
Summary:
  2H2 + 4OH- --> 4H2O + 4e- (anode)
  O2 + 2H2O + 4e- -->4OH- (cathode)
The overall reaction is:
  2H2 + O2 --> 2H2O
                            Ch 23 sec 2
                          Cell Potentials
Electrical Potential - cells ability to
 produce an electric current. Measured
 in Volts.
The electrical potential results from a
 competition for electrons.
E0 = reduction potential
Cell Potential
  E0 = E0red - E0Oxid
Standard Cell Potential
measured potential when the ion
 concentrations in the half-cells are 1M,
 gases are at 101 kPa and temperature
 is 25oC
Half-cell potentials cannot be
 measured. So, the 1/2 cells are
 compared to assigning a hydrogen cell
 0.00 V where reduction of Hydrogen
 occurs.
Reduction takes place at the cathode
Oxidation takes place at the anode
Calculating 1/2 cell potential examples:
  2 examples
Zinc and Hydrogen half cells.
A voltmeter reads +0.76 V.
The zinc is oxidized (anode)
Hydrogen Ions are reduced (cathode)
oxidation Zn --> Zn2+ + 2e-
reduction 2H+ + 2e- --> H2
cell reaction Zn + 2H+ --> Zn2+ + H2
therefore,
  E0 = E0red - E0Oxid
  E0 = E0H+ - E0Zn2+
  +0.76V = 0.00V - E0Zn2+
  E0Zn2+ = -0.76V
The value is negative because the tendency
 for zinc ions to be reduced to zinc metal is
 less than the tendency of hydrogen ions to be
 reduced to hydrogen gas. Therefore it is
 negative because
 Zinc is oxidized.
Second example this time for copper.
Reduction Cu2+ + 2e- --> Cu
Oxidation H2 --> 2H+ + 2e-
Cell reaction Cu2+ + H2 --> Cu + 2H+
Cell potential measured at +0.34V.
So, E0 = E0red - E0Oxid
  +0.34V = E0Cu2+ - 0.00H+
  E0Cu2+ = +0.34V
The potential for Cu ions to be reduced is
 higher than the potential for H ions to be
 reduced. Therefore it is positive because
Copper is reduced.
Table 23.2 in your book on page 688 has a
 list of all the 1/2 cell potentials.
  Calculating Standard Cell Potentials
To function a cell must be constructed
 of 2 half-cells.
If the cell potential for a given redox
 reaction is positive then the reaction is
 spontaneous.
  Calculating Standard Cell Potentials
Example
Determine the cell reaction, the
 standard cell potential, and the half-cell
 that acts as the cathode for a voltaic
 cell composed of the following half-
 cells.
Fe3+ + e- --> Fe2+ E0Fe3+ = +0.77V
Ni2+ + 2e- --> Ni    E0Ni2+ = -0.25V
Answer:
  Reduction takes place in the Fe3+ half cell.
   So, this cell is the Cathode
The 1/2 reactions are:
  Ni --> Ni2+ + 2e-
  2[Fe3+ + e- --> Fe2+]
The cell reaction is:
Ni + 2Fe3+ --> Ni2+ + 2Fe2+

Cell potential is:
E0 = E0red - E0Oxid
       +0.77V - (-0.25V) = +1.02V
      The reaction is spontaneous.
Your Turn:
 Calculate E0cell to determine whether
 the following redox reaction is
 spontaneous as written:
  Ni + Fe2+ --> Ni2+ + Fe
  From table 23.2 in your text
  E0Ni2+ = -0.25V     E0Fe2+ = -0.44V
answer
Nickel is oxidized Iron is reduced
E0= -0.44V - (-0.25V) = -0.19
Reaction will proceed in the reverse
Electrolytic Cells 23.3
                            Electrolytic Cells
                                         23.3
Electrolysis - making a nonspontaneous
 reaction go.
ex - silverplated dishes and utensils, gold-
 plated jewelry and chrome-plated auto parts
The electrolytic cell is an electrochemical cell
 used to cause a chemical change through the
 application of electrical energy (DC Current).
Similarities between voltaic and electric cells:
electrons flow from the anode to the cathode
reduction occurs at cathode
oxidation anode
Differences
Voltaic is spontaneous
Electrolytic is result of outside push of
 electrons
Cathode is negative electrode
 (connected to negative electrode of
 battery) and vice versa for anode.
Electrolysis of water produces hydrogen and
 oxygen gas.
An electrolyte is often used to help conduct
 an electric current. The electrolyte is usually
 redoxed.
The electrolysis of Brine (salt water) produces
 Cl2 + H2 + NaOH
Electrolysis of molten NaCl produces:
liquid Na used in sodium vapor lamps and as
 the coolant in some nuclear reactors.
chlorine gas used to sterilize drinking water
 and the manufacturing of PVC and pesticides.
Which one comes off of which electrode?
sodium gains an electron - cathode
                       Electroplating
The depositing of a thin layer of a metal
 on an object
Common metal used to plate:
  Ag, Au,Cu, Ni, Cr
An object to be silver plated is made
 the cathode. Why?
Anode is the silver to be deposited
Electrolyte is a silver salt.
Also used to purify metals and various
 other processes you will read about.

				
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